7. arithmetic & number theoretic recreations a. Fibonacci numbers

Prob. 25, pp. 20 & 79. Same as Dudeney's AM 82.

Prob. 87: Three by three -- part two, pp. 61 & 120-121. See 7.AC.4 for the distribution of the nine positive digits as ABC, DEF, GHI with the lowest product, which is uniquely given by 147 x 258 x 369 = 13994694.

David Singmaster. Work of 30 Jul 1998 in response to a letter of J. I. Collings. There are 99 solutions of AB*CDE = FG*HIJ, with A < F. Of these, 35 have a leading zero. Trailing zeroes lead to 13 pairs of related solutions, e.g. 23 * 760  =  76 * 230  =  95 * 184. The largest value of the common product is 58560  =  64 * 915 = 80 * 732, as given by Dudeney, AM, prob. 82. The smallest common product is 3588 = 04 * 897 = 23 * 156, while the smallest without a leading zero is 8262 = 18 * 459 = 27 * 306. There are two cases with the same common product and further one of the two products is the same: 18 * 465  =  30 * 279  =  45 * 186. Collings notes that there is only one solution with E or J being three and no leading zeroes
7.AC.5. PAN DIGITAL FRACTIONS
Mittenzwey. 1880. Prob. 141 pp. 30 & 79; 1895?: 161, pp. 33 & 82; 1917: 161, pp. 31 & 79. Use 1, 2, ..., 9 to form three fractions which add to one. Solutions: 9/12 + 5/34 + 7/68 (cf Yoshigahara, below); 21/84 + 9/63 + 5/7; 21/48 + 7/36 + 5/9; 19/76 + 4/32 + 5/8, etc. [Are there really others?]

Pearson. 1907. Part II: Juggling with the digits, pp. 40 41. Examples of ABCD/EFGHI =  1/n for n = 3, 4, ..., 9.

Dudeney. AM. 1917.

Prob. 88: Digital division, pp. 16 & 158. Gives 13458/6729 = 2. Find solutions of ABCDE/FGHI = n for n = 3, 4, ..., 9. Also find the smallest solutions in each case -- e.g. 14658/7329 = 2 is a larger solution than the first example.

Prob. 90: The century puzzle, pp. 16 & 158-159. Write a mixed number, using the 9 positive digits, equal to 100, e.g. 91  5742/638. Says Lucas found 7 ways, but he has shown that there are just 11 ways. One of these has a single digit integer part -- find it. Answer gives all 11 solutions.

Prob. 91: More mixed fractions, pp. 16 & 159. Says he has tried the same question with 100 replaced by other values and gives 12 values to try. However, two of these are impossible. He has found solutions for all values from 1 to 100, except that 1, 2, 3, 4, 15, 18 are impossible, though 15 and 18 can easily be expressed if the integer part is permitted to be zero or if compound fractions are permitted, e.g. 3 (8952/746)/1.

Prob. 92: Digital square numbers, pp. 16 & 159. Find largest and smallest squares using all 9 positive digits. Answers: 923,187,456 = 30384²; 139,854,276  =  11826².

Peano. Giochi. 1924. Prob. 35, p. 10. Gives 6 solutions of 9  =  ABCDE/FGHIJ, three of which have F = 0.

Haldeman-Julius. 1937. No. 112: Half problem, pp. 18 & 26. Use the nine positive digits to make 1/2. Answer is 7293/14586, which is similar to Dudeney's prob. 88.

M. Adams. Puzzle Book. 1939. Prob. B.83: Figure juggling, part 2, pp. 78 & 107. Asks for an example of using the nine positive digits to make 1/2 and remarks that solutions exist for 1/n with n = 3, ..., 9. Gives same solution as Haldeman-Julius.

George S. Terry. The Dozen System. Longmans, Green & Co., NY, 1941. ??NYS -- quoted in Underwood Dudley; Mathematical Cranks; MAA, 1992, p. 25. Express unity as a sum of two fractions which contain all the digits once only, duodecimally. E.g. 136/270 + 48χ/95ε = 1 (χ = 10, ε = 11). Says about five dozen. Terry (or Dudley) says the decimal answer is about one dozen.

Ripley's Believe It or Not, 24th series. Pocket Books, NY, 1975. P. 76. 3/6 = 7/14  =  29/58 uses all nine positive digits. [Are there other examples or other forms??]

Michael Holt. Math Puzzles and Games. Walker Publishing Co., NY, (1977), PB ed., 1983. 9 in ten digits, pp. 26 & 98. ABCDE/FGHIJ  =  9 has six solutions, all given.

James W. Carroll. Letter: Computerizing Sam Loyd. Games 7:5 (May 1983) 6. ABCD/EFGHI = 1/n for n = 2, 3, ..., 9 has 12, 2, 4, 12, 3, 7, 46, 3 solutions.

Nob Yoshigahara. Puzzle problem used on his TV(?) program in Japan and communicated to me at 13th International Puzzle Party, August, 1993. Use the nine positive digits to make A/BC + D/EF + G/HI = 1. There is a unique solution: 5/34 + 7/68 + 9/12.

7.AC.6. OTHER PAN DIGITAL AND SIMILAR PROBLEMS
See also 7.I and 7.I.1 for related problems.
The Family Friend (1856) 149 & 180. Enigmas, Charades, &c. 87 Mathematical Puzzle. "Take all the figures, (i.e., 1 2 3 4 5 6 7 8 9 0,) and place them in such a mode, that, when they are added up, they may be equal to 100." Signed S. W. S. Answer is 76 + 3 10/5 + 8 + 9 4/2.

Magician's Own Book. 1857. The united digits, p. 246. "Arrange the figures 1 to 9 in such order that, by adding them together, they amount to 100." 15 + 36 + 47  =  98 + 2  =  100. = Book of 500 Puzzles, 1859, p. 60. = Boy's Own Conjuring Book, 1860, p. 216. c= Parlour Games for Everybody. John Leng, Dundee & London, nd [1903 -- BLC], p. 41.

Leske. Illustriertes Spielbuch für Mädchen. 1864?

Prob. 564-16, pp. 253 & 395. Combine 1, 2, ..., 9 to make 100. Answer: 1 + 3/6 + 27/54 + 98.

Prob. 564-19, pp. 253 & 395. Add the nine digits to make 100. Answer: 75 9/18  +  24 3/6.

Boy's Own Book. The united digits. 1868: 429. "The figures 1 to 9 may be placed in such order that the whole added together make exactly 100. Thus -- 15 + 36 + 47  =  98 + 2  =  100."

Hanky Panky. 1872. The century of cards, p. 294. 15 + 36 + 47 = 98 + 2 = 100 given with cards.

Mittenzwey. 1880.

Prob. 138, pp. 29-30 & 79; 1895?: 158, pp. 33 & 82; 1917: 158, pp. 31 & 79. Use 1, 2, ..., 9 to make 100. He gives: 95 + 3 + 1 + 6/7 + 4/28; 75 9/18  +  24 3/6 (cf Leske). He says there are many such and many variations, e.g. allowing one repetition and making 1000. [There are ways to make 1000 without any repetitions, e.g. 987 + 6 + 5 + 4 + 2 - 3 - 1.]

Prob. 140 pp. 30 & 79; 1895?: 160, pp. 33 & 82; 1917: 160, pp. 31 & 79. Use 0, 1, ..., 9 to make 10. Solution: 1 35/70 + 8 46/92.

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Card Puzzles No. 4: Century addition, p. 4. Use 1, ..., 9 to add up to 100. Solutions are: 74 + 12 + 3  =  89 + 6 + 5  =  100; same as Book of 500 Puzzles; 19 + 28 + 6 = 53 + 47 = 100.

Anonymous problem proposal with solutions by K. K.; R. Ichikawa; S. Tamano and two papers by T. Hayashi. J. of the Physics School in Tokyo 5 (1896) 82, 99-103, 153-156 & 266-267, ??NYS Abstracted in: Yoshio Mikami, ed.; Mathematical Papers from the Far East; AGM 28 (1910) 16-20; as: A queer number. In base b, we have [12...b * (b 2)] + b-1 = b...21. [I wonder about other solutions of a * x + b = y, where a, b are digits and x, y are pandigital expressions (either with the 9 positive digits or all 10 digits, either separately or together) or y is the reversal of x, etc.]

H. D. Northrop. Popular Pastimes. 1901. No. 17: Magical addition, pp. 69 & 74. "Arrange the figures 1 to 9, so that by adding them together they will make 100. How can this be done?" Solution is: 15 + 36 + 47  =  98 + 2  =  100. Cf Magician's Own Book.

T. Hayashi. On the examination of perfect squares among numbers formed by the arrangements of the nine effective figures. J. of the Physics School in Tokyo 5 (1896) 203-206, ??NYS. Abstracted in: Yoshio Mikami, ed.; Mathematical Papers from the Far East; AGM 28 (1910) 23-25. Says Artemas Martin asked which squares contain all nine positive digits once each and that Biddle found 29 of these. (Cites same J. 5 (1896) 171, ??NYS, for the solutions.) [What about with all 10 digits?].

Clark. Mental Nuts. 1897, no. 98; 1904, no. 42; 1916, no. 44. One in addition. "Place the figures 1 2 3 4 5 6 7 8 9 0 to add 100." Answer: 50 1/2 + 49 38/76.

Ball. MRE, 4th ed., 1905. P. 14. Use the 10 digits to total 1 -- a solution is 35/70 + 148/296 -- or to total 100 -- a solution is 50 + 49 + 1/2 + 38/76. Use the 9 digits to make four numbers which total 100 -- a solution is 78 + 15 + ²9 + ³64.

Ball. MRE, 5th ed., 1911. Pp. 13-14. Briefly restates the material in the 4th ed. as "questions which have been propounded in recent years. ... To the making of such questions of this kind there is no limit, but their solution involves little or no mathematical skill."

M. Adams. Indoor Games. 1912. A clever arrangement, p. 353. Same as Boy's Own Book.

Ball. MRE, 6th ed., 1914. Pp. 13-14. Restates the material in the 5th ed. as "... numerous empirical problems, .... To the making of such questions there is no limit, but their solution involves little or no mathematical skill."

He then introduces the "Four Digits Problem". "I suggest the following problem as being more interesting." Using the digits 1, 2, ..., n, express the integers from 1 up using four different digits and the operations of sum, product, positive integral power and base-10 (or also allowing iterated square roots and factorials). With n = 4, he can get to 88 or to 264. With n = 5, he can get to 231 or 790. Using 0, 1, 2, 3, he can get to 36 (or 40).

Dudeney. AM. 1917. Prob. 13: A new money puzzle, pp. 2-3 & 148-149. States largest amount of old English money expressible with all nine positive digits is £98765  4s  3½d. Asks for the smallest amount. Answer: £2567 18s 9¾d.

Ball. MRE, 9th ed., 1920. Pp. 13-14. In the "Four Digits Problem", he considers n = 4, i.e. using 1, 2, 3, 4, and discusses the operations in more detail. Using sum, product, positive integral power and base-10 notation, he can get to 88. Allowing also finitely iterated square roots and factorials, he can get to 264. Allowing also negative integral indices, he can get to 276. Allowing also fractional indices, he can get to 312. He then mentions using 0, 1, 2, 3 or four of the five digits 1, ..., 5.

Ball. MRE, 10th ed., 1922. Pp. 13-14. In the "Four Digits Problem", he repeats the material of the 9th ed., but at the end he adds that using all of the five digits, 1, ..., 5, he has gotten to 3832 or 4282, depending on whether negative and fractional indices are excluded or allowed.

Hummerston. Fun, Mirth & Mystery. 1924. Century making, p. 66.

15 + 36 + 47 = 98 + 2 = 100.

Wood. Oddities. 1927.